(log base a of 25)/(log base a of 3a)

Question

anonymous · Answer

$$\log_{a}25 \div \log_{a}  3a$$

anonymous · Answer

Oops,sorry I forgot to add this $$\log_{a} 3 =0.477 ~and ~ \log_{a} 5 =0.699$$

anonymous · Answer

Note that $\large \dfrac{\log_a 25}{\log_a(3a)} = \dfrac{\log_a(5^2)}{\log_a3 + \log_a a}=\ldots$

Can you take things from here? :-)

anonymous · Answer

Yup,but what if I were to apply product law for the numerator?

anonymous · Answer

You could do that as well by letting $\large 25 = 5\cdot 5$.  

Then $\large \log_a25 = \log_a(5\cdot 5) = \log_a5+\log_a5 = 2\log_a5 = \log_a(5^2)$.

You can now see that both ways are equivalent.

I hope this clarifies things! :-)

anonymous · Answer

Yes,it does,Thanks

anonymous · Answer

Um,could you give me the final answer just to check?

anonymous · Answer

You should end up with $\large \dfrac{2\log_a 5}{\log_a3 + 1} = \dfrac{2\cdot 0.699}{0.477+1} \approx 0.946513 $

anonymous · Answer

Ahh,now it makes sense.I completely forgot about the plus one.lol thanks